Respuesta :
Answer:
[tex]p_v =P(Z>4.146)=0.0000169[/tex]
Based on the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .
Step-by-step explanation:
1) Data given and notation
n=170 represent the random sample taken
X=58 represent the student's who belief that such software unfairly targets students
[tex]\hat px=\frac{58}{170}=0.341[/tex] estimated proportion of students who belief that such software unfairly targets students
[tex]\hat p=\frac{112}{170}=0.659[/tex] estimated proportion of students who NOT belief that such software unfairly targets students
[tex]p_o=0.50[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level (no given)
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
p= proportion of student's who belief that such software unfairly targets students
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that that a majority of students at the university do not share this belief. :
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p>0.5[/tex]
We assume that the proportion follows a normal distribution.
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly (different,higher or less) from a hypothesized value [tex]p_o[/tex].
Check for the assumptions that he sample must satisfy in order to apply the test
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
[tex]np_o =170*0.5=85>10[/tex]
[tex]n(1-p_o)=170*(1-0.5)=85>10[/tex]
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{170}}}=4.146[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a one right side test the p value would be:
[tex]p_v =P(Z>4.146)=0.0000169[/tex]
Based on the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students who NOT belief that such software unfairly targets students is higher than 0.5 or 50% .
